Question

1. Fie ΔABC si punctul M ∈ [AB] a.i. AM=$\frac{3}{5}*MB$. Exprimati vectorul CM in functie de vectorii CA si CB.

Answer 1

In ΔABC, exprimat cu vectori, suma a doi vectori dupa regula triunghiului, este: AB= AC+CB, in relatia data, aplicand proportii derivate obtinem :
$\frac{AM}{MB}= \frac{3}{5},rezulta: \frac{AM}{AM+MB}= \frac{3}{3+5},adica: \frac{AM}{AB}= \frac{3}{8},sauAM= \frac{3}{8}AB=$$\frac{3}{8}(AC+CB)$. In ΔACM, avem: CM=CA+AM=CA+(3/8)(AC+CB)=
CA+(3/8)(-CA+CB)=(5/8)CA+(3/8)CB=(1/8)(5CA+3CB), ceea ce sa cerut.